- #1

Euge

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- #1

Euge

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- #2

anuttarasammyak

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with no reard to individual value of x. Summing up this in x interval of [0,x] x>0

[tex]|f(x)-f(0)|\leq \frac{\epsilon}{\delta}|x|+\epsilon[/tex]

we get same formula for [x,0] x<0. So for any x

[tex]|f(x)| \leq \frac{\epsilon}{\delta}|x|+|f(0)|+\epsilon \leq A+B|x|[/tex]

where [tex]|f(0)|+\epsilon \leq A , \frac{\epsilon}{\delta}\leq B[/tex]

[EDIT]

Thanks to the sugegstion in post #13, I would add a sentense to the second line :

with no reard to individual value of x.

Last edited:

- #3

bob012345

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Isn't this just the definition of a uniformly continuous function?

- #4

Office_Shredder

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Isn't this just the definition of a uniformly continuous function?

I think the definition of uniformly continuous is a local definition - for any ##\epsilon## there exists ##\delta## such that ##|f(y)-f(x)|<\epsilon## if ##|x-y|<\delta##. The problem statement here is a statement about the whole function, not just two points that are very close together.

You have to do at least some work to get from one to the other.

- #5

PeterDonis

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I thought just plain old continuous was the local definition, and uniformly continuous was the global definition.I think the definition of uniformly continuous is a local definition

Not really; it's a property that must be satisfied at all values of ##x##, but so is ordinary continuity. But the definition of the property, for any given value of ##x##, only involves that value of ##x##; it does not involve the whole set of values of ##x## all at once.The problem statement here is a statement about the whole function

- #6

PeterDonis

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I don't think so, because a uniformly continuous function does not have to be linear. The problem seems to be asking you to show that any uniformly continuous function can be approximated by a linear function.Isn't this just the definition of a uniformly continuous function?

- #7

PeterDonis

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https://people.math.wisc.edu/~jwrobbin/521dir/cont.pdf

A quick summary: if we look at epsilons and deltas, continuous means epsilon can vary, and delta can vary with both epsilon and ##x##; uniformly continuous means that epsilon can vary, and delta can vary with epsilon but not with ##x##; and the problem statement in the OP requires that there is a single epsilon and delta that work for all values of ##x##.

- #8

Office_Shredder

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Once you pick a choice of ##\epsilon## uniform continuity only says things about points that are close together. The fact that the same Delta and epsilon work is global, but the definition still says if two inputs are close together, so are the outputs.I thought just plain old continuous was the local definition, and uniformly continuous was the global definition.

Not really; it's a property that must be satisfied at all values of ##x##, but so is ordinary continuity. But the definition of the property, for any given value of ##x##, only involves that value of ##x##; it does not involve the whole set of values of ##x## all at once.

- #9

PeterDonis

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This is just as true of the problem statement in the OP. So I don't see that there is a useful "local" vs. "global" distinction here.Once you pick a choice of ##\epsilon## uniform continuity only says things about points that are close together. The fact that the same Delta and epsilon work is global, but the definition still says if two inputs are close together, so are the outputs.

- #10

Office_Shredder

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No, the Op immediately gives a way to compare the output of two inputs no matter how far away they are. The definition only let's you compare the output of two inputs that are close together.

@anuttarasammyak's proof demonstrates this, they take something which is only true on a small interval, and they extrapolate it to demonstrate something that is true across any arbitrary interval. It's a bit subtle but it's there.

@anuttarasammyak's proof demonstrates this, they take something which is only true on a small interval, and they extrapolate it to demonstrate something that is true across any arbitrary interval. It's a bit subtle but it's there.

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- #11

PeterDonis

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The OP statement doesn't compare the function's values at two different values of ##x## at all.the Op immediately gives a way to compare the output of two inputs no matter how far away they are

- #12

PeterDonis

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I'm not sure this proof is valid.@anuttarasammyak's proof

- #13

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It's conceptually a bit messy, IMO, as it leaves ##\epsilon## arbitrary. He should have let ##\epsilon =1##, for example.I'm not sure this proof is valid.

Other than that, the idea is sound.

- #14

BWV

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What do you all think of ChatGPT's answer below?

$$|f(x) - f(y)| < 1.$$

Take any ##x \in \mathbb{R}##. Then, for any ##y \in \mathbb{R}## with ##|y| \le |x| + \delta##, we have

$$|f(x) - f(y)| < 1.$$

In particular, if we take ##y = 0##, we get

$$|f(x)| \le 1 + |f(0)|.$$

Therefore, we can take ##A = |f(0)|## and ##B = 1##. This shows that, for some positive constants ##A## and ##B##, we have

$$|f(x)| \le A + B|x|$$

for all ##x \in \mathbb{R}##.

Note that the choice of ##A## and ##B## is not unique: any constants ##A'## and ##B'## such that ##A' \ge A## and ##B' \ge B## will also work. For example, we can take ##A' = \max {|f(x)| : x \in \mathbb{R}}## and ##B' = \delta^{-1}##, where ##\delta## is the constant from the uniform continuity of ##f##.

- #15

Office_Shredder

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It's got the right shape, but serious errors.

- #16

WWGD

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Bounded derivative is sufficient, per the mvt ( though function does not have to be differentiable to be UC.I don't think so, because a uniformly continuous function does not have to be linear. The problem seems to be asking you to show that any uniformly continuous function can be approximated by a linear function.

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